Answer
The Taylor series centered at $c=5$ is
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{4^{n + 1}}}}{\left( {x - 5} \right)^n}$
It is valid for $\left| {x - 5} \right| \lt 4$.
Work Step by Step
We compute the derivatives and list them in the table below:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}&{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( 5 \right)}}{{n!}}}\\
0&{\dfrac{1}{{1 - x}}}&{\dfrac{1}{{1 - x}}}&{ - \dfrac{1}{4}}\\
1&{\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}}&{\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}}&{\dfrac{1}{{16}}}\\
2&{\dfrac{2}{{{{\left( {1 - x} \right)}^3}}}}&{\dfrac{1}{{{{\left( {1 - x} \right)}^3}}}}&{ - \dfrac{1}{{64}}}\\
3&{\dfrac{6}{{{{\left( {1 - x} \right)}^4}}}}&{\dfrac{1}{{{{\left( {1 - x} \right)}^4}}}}&{\dfrac{1}{{256}}}\\
4&{\dfrac{{24}}{{{{\left( {1 - x} \right)}^5}}}}&{\dfrac{1}{{{{\left( {1 - x} \right)}^5}}}}&{ - \dfrac{1}{{1024}}}
\end{array}$
By Theorem 1, the Taylor series is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 5 \right)}}{{n!}}{\left( {x - 5} \right)^n}$
Using the table above, we get
$T\left( x \right) = - \dfrac{1}{4} + \dfrac{1}{{16}}\left( {x - 5} \right) - \dfrac{1}{{64}}{\left( {x - 5} \right)^2} + \dfrac{1}{{256}}{\left( {x - 5} \right)^3} - \dfrac{1}{{1024}}{\left( {x - 5} \right)^4} + \cdot\cdot\cdot$
So, $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{4^{n + 1}}}}{\left( {x - 5} \right)^n}$.
Let ${a_n} = \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{4^{n + 1}}}}{\left( {x - 5} \right)^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{4^{n + 1}}}}{{{4^{n + 2}}}}\dfrac{{{{\left( {x - 5} \right)}^{n + 1}}}}{{{{\left( {x - 5} \right)}^n}}}} \right| = \dfrac{1}{4}\left| {x - 5} \right|$
We find that $\rho \lt 1$ if $\dfrac{1}{4}\left| {x - 5} \right| \lt 1$, or $\left| {x - 5} \right| \lt 4$.
Thus, the Taylor series centered at $c=5$ is $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{4^{n + 1}}}}{\left( {x - 5} \right)^n}$. It is valid for $\left| {x - 5} \right| \lt 4$.
